Survival Analysis

Dr. F.J. Rodenburg
June 2024 (rev. 1)

Suppose...

A new cancer treatment is compared to an existing alternative.

Primary outcome: Patient survival

plot of chunk unnamed-chunk-2

plot of chunk unnamed-chunk-3

Kaplan-Meijer

Example table of patients:

patient	time	status	age	sex
1	5	dead	65	female
2	11	dead	74	male
3	11	dead	81	male
4	11	dead	67	male
5	12	dead	74	male
…	…	…	…	…

\begin{aligned} \hat{S} (day 5) & = 1 \cdot \frac{227}{228} \approx 99.6 % \end{aligned}

$\begin{aligned} \hat{S}(\text{day 5}) &= 1 \cdot \frac{227}{228} \approx 99.6\% \\ \\ \\ \\ \end{aligned}$

Kaplan-Meijer

Example table of patients:

patient	time	status	age	sex
1	5	dead	65	female
2	11	dead	74	male
3	11	dead	81	male
4	11	dead	67	male
5	12	dead	74	male
…	…	…	…	…

\begin{aligned} \hat{S} (day 5) & = 1 \cdot \frac{227}{228} \approx 99.6 % \\ \hat{S} (day 11) & = 1 \cdot \frac{227}{228} \cdot \frac{224}{227} \approx 98.2 % \end{aligned}

$\begin{aligned} \hat{S}(\text{day 5}) &= 1 \cdot \frac{227}{228} \approx 99.6\% \\ \\ \hat{S}(\text{day 11}) &= 1 \cdot \frac{227}{228} \cdot \frac{224}{227} \approx 98.2\% \\ \\ \end{aligned}$

Kaplan-Meijer

Example table of patients:

patient	time	status	age	sex
1	5	dead	65	female
2	11	dead	74	male
3	11	dead	81	male
4	11	dead	67	male
5	12	dead	74	male
…	…	…	…	…

\begin{aligned} \hat{S} (day 5) & = 1 \cdot \frac{227}{228} \approx 99.6 % \\ \hat{S} (day 11) & = 1 \cdot \frac{227}{228} \cdot \frac{224}{227} \approx 98.2 % \\ \hat{S} (day 12) & = 1 \cdot \frac{227}{228} \cdot \frac{224}{227} \cdot \frac{223}{224} \approx 97.8 % \end{aligned}

$\begin{aligned} \hat{S}(\text{day 5}) &= 1 \cdot \frac{227}{228} \approx 99.6\% \\ \\ \hat{S}(\text{day 11}) &= 1 \cdot \frac{227}{228} \cdot \frac{224}{227} \approx 98.2\% \\ \\ \hat{S}(\text{day 12}) &= 1 \cdot \frac{227}{228} \cdot \frac{224}{227} \cdot \frac{223}{224} \approx 97.8\% \end{aligned}$

Kaplan-Meijer

For each time point $t_i$ , on which at least one event occurred, until the last time point $t_j$ :

\hat{S} (t) = \prod_{i = 1}^{j} (1 - \frac{d_{i}}{n_{i}})

$\hat{S}(t) = \prod_{i = 1}^j \bigg( 1 - \frac{d_i}{n_i} \bigg)$

$d_i$ is the number of events at time point $t_i$ ,

$n_i$ is the number at risk at time point $t_i$ .

Compare Survival Curves

Using a log-rank test:
- Observed vs expected number of events at each time point;
- Much like an ordinary $\chi^2$ -test;
Using a more general framework (Cox proportional hazards).

plot of chunk unnamed-chunk-42

Summary

Survival analysis is used to study time until event data.

Censoring and truncation;

Censored data can be accounted for using the Kaplan–Meijer estimator;

Survival curves can be compared with the log-rank test (a

χ^{2}

$\chi^2$ -test);

Regression analysis can be done with Cox proportional hazards;

Censored and truncated data are possible in all kinds of research.

Summary

Video: https://youtu.be/t6vdjwhauF8
Example from the video:

library("survival")
require("ggsurvfit")

# Fit a Kaplan-Meijer curve
KM <- survfit2(Surv(time, status) ~ sex, data = lung)                         

# Plot the result
ggsurvfit(KM) + labs(x = "Days", y = "Probability of survival") +
  add_risktable() + add_confidence_interval() + theme(panel.grid.major = element_blank(), panel.grid.minor = element_blank(),
panel.background = element_blank(), axis.line = element_line(colour = "black"))

# Perform a log-rank test
survdiff(Surv(time, status) ~ sex, data = lung)

# Estimate survival at a given time (days)
summary(KM, times = 250)